Entropy on Riemann Surfaces and the Jacobians of Finite Covers

Entropy on Riemann Surfaces and the Jacobians of Finite Covers

Entropy on Riemann surfaces and the Jacobians of finite covers Curtis T. McMullen 20 June, 2010 Abstract This paper characterizes those pseudo-Anosov mappings whose en- tropy can be detected homologically by taking a limit over finite covers. The proof is via complex-analytic methods. The same methods show the natural map g h, which sends a Riemann surface to the M → A Jacobians of all of its finiteQ covers, is a contraction in most directions. Contents 1 Introduction............................ 1 2 Oddorderzeros.......................... 3 3 Siegelspace............................ 3 4 Teichm¨ullerspace......................... 6 5 Contraction............................ 7 6 Conclusion ............................ 8 A The hyperbolic metric via Jacobians of finite covers . 9 Research supported in part by the NSF. 2000 Mathematics Subject Classification: 32G15, 37E30. Keywords: Entropy, pseudo– Anosov maps, Kobayashi metric, Siegel space, Teichm¨uller space, homology. 1 Introduction Let f : S S be a pseudo-Anosov mapping on a surface of genus g with → n punctures. It is well-known that the topological entropy h(f) is bounded below in terms of the spectral radius of f ∗ : H1(S, C) H1(S, C); we have → log ρ(f ∗) h(f). ≤ If we lift f to a map f : S S on a finite cover of S, then its entropy stays → the same but the spectral radius of the action on homology can increase. e e e We say the entropy of f can be detected homologically if h(f) = sup log ρ(f ∗ : H1(S) H1(S)), → where the supremum is taken over alle finitee covers toe which f lifts. In this paper we will show: Theorem 1.1 The entropy of a pseudo-Anosov mapping f can be detected homologically if and only if the invariant foliations of f have no odd-order singularities in the interior of S. The proof is via complex analysis. Hodge theory provides a natural em- bedding g g from the moduli space of Riemann surfaces into the M → A moduli space of Abelian varieties, sending X to its Jacobian. Any char- acteristic covering map from a surface of genus h to a surface of genus g, branched over n points, provides a similar map g,n h h. (1.1) M → M →A It is known that the hyperbolic metric on a Riemann surface X can be reconstructed using the metrics induced from the Jacobians of its finite covers ([Kaz]; see the Appendix). Similarly, it is natural to ask if the Te- ichm¨uller metric on g,n can be recovered from the Kobayashi metric on M h, by taking the limit over all characteristic covers g,n. We will show such A C a construction is impossible. Theorem 1.2 The natural map g,n h is not an isometry for M → Cg,n A the Kobayashi metric, unless dim g,n = 1Q. M It is an open problem to determine if the Kobayashi and Carath´eodory met- rics on moduli space coincide when dim g,n > 1 (see e.g. [FM, Prob 5.1]). M An equivalent problem is to determine if Teichm¨uller space embeds holo- morphically and isometrically into a (possibly infinite) product of bounded 1 symmetric domains. Theorem 1.2 provides some support for a negative an- swer to this question. Here is a more precise version of Theorem 1.2, stated in terms of the lifted map J g,n h Hh T →T → from Teichm¨uller space to Siegel space determined by a finite cover. Theorem 1.3 Suppose the Teichm¨uller mapping between a pair of distinct points X,Y g,n comes from a quadratic differential with an odd order ∈ T zero. Then sup d(J(X), J(Y )) < d(X,Y ), where the supremum is taken overe alle compatible finite covers of X and Y . Conversely, if the Teichm¨uller map from X to Y has only even order singu- larities, then there is a double cover such that d(J(X), J(Y )) = d(X,Y ) (cf. [Kra]). In particular, the complex geodesics generated by squares of holo- e e morphic 1-forms map isometrically into g. The only directions contracted A by the map g h are those identified by Theorem 1.3. M → A Theorem 1.1 followsQ from Theorem 1.3 by taking X and Y to be points on the Teichm¨uller geodesic stabilized by the mapping-class f. It would be interesting to find a direct topological proof of Theorem 1.1. As a sample application, let β Bn be a pseudo-Anosov braid whose ∈ monodromy map f : S S (on the n-times punctured plane) has an odd → order singularity. Then Theorem 1.1 implies the image of β under the Burau representation satisfies log sup ρ(B(q)) < h(f). |q|=1 Indeed, ρ(B(q)) at any d-th root of unit is bounded by ρ(f ∗) on a Z/d cover S [Mc2]. This improves a result in [BB]. Similar statements hold for other e homological representations of the mapping–class group. Notes and references. For C∞ diffeomorphisms of a compact smooth manifold, one has h(f) log sup ρ(f ∗ Hi(X)) [Ym], and equality holds ≥ i | for holomorphic maps on K¨ahler manifolds [Gr]. The lower bound h(f) ≥ log ρ(f ∗ H1(X)) also holds for homeomorphisms [Mn]. For more on pseudo- | Anosov mappings, see e.g. [FLP], [Bers] and [Th]. A proof that the inclusion of g,n into universal Teichm¨uller space is a T contraction, based on related ideas, appears in [Mc1]. 2 2 Odd order zeros We begin with an analytic result, which describes how well a monomial zk of odd order can be approximated by the square of an analytic function. Theorem 2.1 Let k 1 be odd, and let f(z) be a holomorphic function on ≥ the unit disk ∆ such that f(z) 2 = 1. Then | | R k 2 z √k + 1√k + 3 f(z) Ck = < 1. Z∆ z ≤ k + 2 | | Here the integral is taken with respect to Lebesgue measure on the unit disk. n Proof. Consider the orthonormal basis en(z) = anz , n 0, an = 2 ≥ √n + 1/√π, for the Bergman space Lα(∆) of analytic functions on the disk with f 2 = f(z) 2 < . With respect to this basis, the nonzero entries k k2 | | ∞ in the matrixR of the symmetric bilinear form Z(f, g)= f(z)g(z)zk/ z k are | | given by R k 2√n + 1√k n + 1 Z(en, ek−n)= anak−n z = − Z∆ | | k + 2 · In particular, Z(ei, ei) = 0 for all i (since k is odd), and Z(ei, ej) = 0 for all i, j > k. Note that the ratio above is less than one, by the inequality between the arithmetic and geometric means, and it is maximized when n< k/2 < n + 1. Thus the maximum of Z(f,f) / f 2 over L2 (∆) is achieved when | | k k α f = en + en , n = (k 1)/2, at which point it is given by Ck. +1 − 3 Siegel space In this section we describe the Siegel space of Hodge structures on a surface S, and its Kobayashi metric. Hodge structures. Let S be a closed, smooth, oriented surface of genus g. Then H1(S) = H1(S, C) carries a natural involution C(α) = α fixing H1(S, R), and a natural Hermitian form √ 1 α, β = − α β h i 2 ZS ∧ 3 of signature (g, g). A Hodge structure on H1(S) is given by an orthogonal splitting H1(S)= V 1,0 V 0,1 ⊕ such that V 1,0 is positive-definite and V 0,1 = C(V 1,0). We have a natural norm on V 1,0 given by α 2 = α, α . k k h i The set of all possible Hodge structures forms the Siegel space H(S). To describe this complex symmetric space in more detail, fix a splitting H1(S) = W 1,0 W 0,1. Then for any other Hodge structure V 1,0 V 0,1, ⊕ ⊕ there is a unique operator Z : W 1,0 W 0,1 → such that V 1,0 = (I + Z)(W 1,0). This means V 1,0 coincides with the graph of Z in W 1,0 W 0,1. ⊕ The operator Z is determined uniquely by the associated bilinear form Z(α, β)= α, CZ(β) h i on W 1,0, and the condition that V 1,0 V 0,1 is a Hodge structure translates ⊕ into the conditions: Z(α, β)= Z(β, α) and Z(α, α) < 1 if α = 1. (3.1) | | k k Since the second inequality above is an open condition, the tangent space at the base point p W 1,0 W 0,1 is given by ∼ ⊕ 1,0 1,0 TpH(S)= symmetric bilinear maps Z : W W C . { × → } Comparison maps. Any Hodge structure on H1(S) determines an iso- morphism 1,0 1 R V ∼= H (S, ) (3.2) sending α to Re(α) = (α + C(α))/2. Thus H1(S, R) inherits a norm and a complex structure from V 1,0. Put differently, (3.2) gives a marking of V 1,0 by H1(S, R). By com- posing one marking with the inverse of another, we obtain the real-linear comparison map T = (I + Z)(I + CZ)−1 : W 1,0 V 1,0 (3.3) → between any pair of Hodge structures. It is characterized by Re(α) = Re(T (α)). 4 Symmetric matrices. The classical Siegel domain is given by Hg = Z Mg(C) : Zij = Zji and I ZZ 0 .

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